A B D G H 1 Test No Normal Force Shear Force Shear Stress 2 N N Area Of Sample A 45 X 37 1665 Mm 2 3 77 4 120 5 1 (117.61 KiB) Viewed 10 times
A B D G H 1 Test no. Normal force Shear force Shear stress 2 (N) (N) Area of sample A = 45 x 37 = 1665 mm^2 3 77 4 120.5 T' (N/mm2) 140 0.084084084 171 0.102702703 188 0.112912913 287 0.172372372 385 0.231231231 5 144 Normal stress on = Normal force A 6 281 7 416 For test 1 8 ơ'n = 77/1665 = 0.0462 9 10 Shear stress = shear force/ area 11 12 For test 2 13 T'= 140/1665 = 0.084 14 15 16 a) The equation for Coulomb's failure criteria is given as 17 T'= c + o'n x tang 18 19 This equation represents the equation of a line, between I' as ordinate and as abscissa, having a slope of tang (ø is friction angle) 20 and the line has a intercept of c' i.e, cohesion with the [' axis. 21 22 So plot of shear stress and normal stress has been shown below: 23 24 25 26 Shear stress Vs Normal Stress 27 0.3 28 29 0.25 30 31 0.2 32 0.15 33 34 0.1 35 36 0.05 37 38 0 39 0.1 0.15 40 41 42 12345 0 0.05 Normal stress o'n (N/mm2) 0.046246246 0.072372372 0.086486486 0.168768769 0.24984985 0.2 F 0.25 K L
Worksheet 4 - WEEK 10- Shear strength of soil 2. In the data set provided in previous problem consider the soil was drained and had a pore water pressure of 29.43kPa. ● Plot the graph using Coulomb's failure criteria. Find out the effective cohesion parameter in kPa and effective friction angle for the soil. Did you see any change with problem 1? Please make comment. (Note and hint: pore water pressure only acts vertically, thus there won't be any effect on the horizontal direction or in other words for shear.) (50)
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